Video Lecture
Theory For Making Notes
Introduction
We know when the velocity of a body changes at a constant rate its acceleration is said to be constant. Similarly When the velocity of a body changesat a non uniform rate, that means the velocity changes by different amount in same time interval the acceleration of the body is said to be non uniform or variable. It happens when the acceleration changes with either Time or Position .
Hence there are three types of question based on relations of acceleration
(i) Acceleration given as a Function of Time i.e. a = f(t)
(ii) Acceleration as a Function of position i.e. a = f(x)
(iii) Acceleration given as a Function of Velocity i.e. a = f(v)
Illustration based on (i)
The acceleration of a particle is given by a = 4t – 30, where a is in m/s2 and t is in s. Determine the velocity and displacement as functions of time. The initial displacement at t = 0 is –5m, and the initial velocity is 3 m/s.
Solution
a = \displaystyle \frac{{dv}}{{dt}} = 4t – 30
dv= (4t – 30)dt
\displaystyle \int\limits_{3}^{v}{{dv}}=\int\limits_{0}^{t}{{\left( {4t-30} \right)dt}}
Þ v – 3 = 2t2 – 30 t
or v = 3 – 30t + 2t2 m/s
Now v = \displaystyle \frac{{dx}}{{dt}} = 3 – 30 t + 2t2
dx= (3 – 30t + 2t2)dt
\displaystyle \int\limits_{{-5}}^{x}{{dx}}=\int\limits_{0}^{t}{{\left( {3-30t+2{{t}^{2}}} \right)dt}}
x + 5 = 3t – 15t2 + \frac{2}{3}t 3
or x = -5 + 3t – 15t2 + \frac{2}{3}t 3 m
Illustration based on (ii)
The acceleration of a body is related to its position as a = kx2. If at x=0 the velocity of the body is 4m/sec. Find the relation of velocity with position x.
Solution
Given \displaystyle a=k{{x}^{2}}
hence \displaystyle \frac{{dv}}{{dt}}=k{{x}^{2}}
or \displaystyle \frac{{dv}}{{dx}}.\frac{{dx}}{{dt}}=k{{x}^{2}}
or \displaystyle \frac{{dv}}{{dx}}.v=k{{x}^{2}}
or \displaystyle vdv=k{{x}^{2}}dx
or \displaystyle \int\limits_{4}^{v}{{vdv}}=k\int\limits_{0}^{x}{{{{x}^{2}}dx}}
or \displaystyle \left[ {\frac{{{{v}^{2}}}}{2}} \right]_{4}^{v}=k\left[ {\frac{{{{x}^{3}}}}{3}} \right]_{0}^{x}
or \displaystyle \frac{{{{v}^{2}}}}{2}-\frac{{{{4}^{2}}}}{2}=k\left[ {\frac{{{{x}^{3}}}}{3}-0} \right]
or \displaystyle {{v}^{2}}=2k\left[ {\frac{{{{x}^{3}}}}{3}} \right]+16
or \displaystyle v=\sqrt{{2k\left[ {\frac{{{{x}^{3}}}}{3}} \right]+16}}
Illustration based on (iii)
The acceleration of a body is related to its velocity as a = kv2. If at x=0 the velocity of the body is 5m/sec. Find the relation of velocity with time t.
Solution
given \displaystyle a=k{{v}^{2}}
\displaystyle \Rightarrow \text{ }\frac{{dv}}{{dt}}=k{{v}^{2}}
\displaystyle \Rightarrow \text{ }\frac{{dv}}{{{{v}^{2}}}}=kdt
\displaystyle \Rightarrow \text{ }\int\limits_{5}^{v}{{\frac{{dv}}{{{{v}^{2}}}}=k\int\limits_{0}^{t}{{dt}}}}
\displaystyle \Rightarrow \text{ }\left[ {\frac{{{{v}^{{-2+1}}}}}{{-2+1}}} \right]_{5}^{v}=k\left[ t \right]_{0}^{t}
\displaystyle \Rightarrow \text{ -}\left[ {\frac{1}{v}} \right]_{5}^{v}=kt
\displaystyle \Rightarrow \text{ }\left[ {\frac{1}{v}-\frac{1}{5}} \right]=-kt
\displaystyle \Rightarrow \text{ }\frac{1}{v}=-kt+\frac{1}{5}
\displaystyle \Rightarrow \text{ }v=\frac{5}{{1-5kt}}
Remember that in the motion with variable acceleration, the graphs between different quantities depends on their mutual relation based on the situation given in the questions
Practice Questions (Level-1)
1.
Choose the incorrect statement for non uniformly accelerated motion
(a) the position time graph may be a parabola
(b) the position time graph may be a hyperbola
(c) the velocity time graph may be a parabola
(d) the velocity time graph may be a hyperbola
Ans. (a)
2.
In a non uniform motion with nonuniform acceleration
(a) The velocity may change but the speed may remains constant with time
(b) The acceleration and velocity both may change with time
(c) The acceleration may change with position of the body
(d) All of the above three options are correct
Ans. (d)
3.
The position of a body is given as x = 2t3 -3t2 -12t -2. (where t ≥0) . The acceleration of the body when the velocity is zero must be
(a) 12 m/sec2
(b) 10 m/sec2
(c) 18 m/sec2
(d) 14 m/sec2
Ans . (c)
4.
In the previous question the time at which the velocity-time graph cuts the acceleration-time graph when drawn on the same diagram is
(a) 2.3 sec
(b) 3.3 sec
(c) 3.8 sec
(d) 6.6 sec
Ans. (b)
5.
If the velocity of a body is given as v = 2t2 – 2t +1. And it is given at t=0 , x=4. Find the position of the body when acceleration is zero.
(a) 3/8
(b) 7/8
(c) 2/5
(d) 3/5
Ans. (b)
6.
The position x of a body is given by x = (u-2)t + 2(a+4)t2 + 10 . If at t=0 , velocity of body is 10m/sec, find u
(a) 10
(b) 16
(c) 12
(d) 18
Ans. (c)
7.
In the previous question assuming `a’ to be greater than 0, then the graph between acceleration and time is
(a) A straight line parallel to time axis cutting acceleration axis at `a’
(b) A straight line with positive slope
(c) A parabola symmetrical to time axis
(d) A straight line parallel to time axis
Ans. (d)
8.
Which of the follwing relation does not represent a non uniformly accelerated motion.
x = position , t = time , v = velocity , a = acceleration
(a) a = v.x
(b) x.v = t
(c) v2 = 4x
(d) a.x3 = 1
Ans. (c)
9.
The acceleration of a body is given by a= 2t + 4. If at t=0 the position x=10 and velocity v=5. Then the relation for instantaneous velocity v(t) is given by
(a) t2 + 4t – 5
(b) t2 + 4t + 5
(c) 2t2 + 4t + 5
(d) t2 + 2t + 5
Ans. (b)
10.
In the previous question the relation between position and time is given by
(a) \displaystyle x=\frac{{{{t}^{3}}}}{3}+4{{t}^{2}}+10t+10
(b) \displaystyle x=\frac{{{{t}^{2}}}}{2}+3{{t}}+10
(c) \displaystyle x=\frac{{{{t}^{3}}}}{3}+2{{t}^{2}}+5t+10
(d) none
Ans. (c)
Practice Questions (Level-2)
1.
A body is moving along a straight line according to the law x = 16t – 6t2 where x is measure in metre and t in second. Find the acceleration of the body when x=2m.
(a) 13 ms2
(b) –12 ms2
(c) 12ms2
(d) –14 ms2
Ans. (b)
2.
The relation between time t and distance x is t = a x2 – bx where a and b are constants. What is the magnitude of retardation if v is the velocity at any time t ?
(a) \displaystyle 5\alpha {{v}^{3}}
(b) \displaystyle 3\alpha {{v}^{3}}
(c) \displaystyle 2\alpha {{v}^{3}}
(d) \displaystyle 4\alpha {{v}^{3}}
Ans. (c)
3.
A particle starting from rest, moves in a straight line and its acceleration is given by a = 50 – 36t2 m/s2. Where t is in seconds. Determine the velocity of the particle when it has traveled 52 m and the time taken by it before it comes to rest again. ( given at t=0 , x & v=0 )
(a) 6 m/s; 2.04 second
(b) 4 m/s; 2.04 second
(c) 5 m/s; 3.04 second
(d) None of these
Ans. (b)
4.
The acceleration of a particle is given by the relation a = 90 – 6x2, where a is expressed in cm/sec2 and x in centimeter. If the particle starts with zero initial velocity at position x = 0, determine the velocity when x = 5 cm and the position where velocity is maximum.
(a) 20 cm/s; 2.873 cm
(b) 10 cm/s; 3.873 cm
(c) 20 cm/s; 3.873 cm
(d) 10 cm/s; 3.873 cm
Ans. (c)
5.
During motion of a body in a viscous medium the acceleration is given by \displaystyle a = A – Bv, where A and B are constants. The terminal velocity of the body is
(a) B/A
(b) B + A
(c) A/B
(d) A.B
Ans. (c)
6.
In a straight line motion the acceleration of a body is given as \displaystyle a\,=\,-\cos \,\,t\,\,,\,\,\,at\,\,\,t\,\,=0\,\,,\,\,\,\,u\,=\,0,\ and\,\,x\,\,=\,\,1. Find distance traveled by the body from t = 0 to t = 2p.
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (d)
7.
A particle moves along a straight line so that its velocity at time t is given by \displaystyle v\left( t \right)={{t}^{2}}-t-6 (measured in meters per second). Find the displacement of the particle during the time period \displaystyle 1\le \,t\,\le \,4 and the distance traveled during this time period.
(a) 4.5 m, 10.17 m
(b) 5.5 m, 10.20 m
(c) 3.5 m, 9.20 m
(d) 5.6 m, 11.20 m
Ans. (a)
8.
Find the position function [ x(t) ] of a particle that moves with velocity \displaystyle v\,\left( t \right)=\cos \,\,\pi t along a straight line, assuming that at t = 0 , x = 4 .
(a) \displaystyle x\left( t \right)\,=\,\frac{1}{\pi }\,\,\sin \,\,\pi \,t\,-\,4
(b) \displaystyle x\left( t \right)\,=-\,\frac{1}{\pi }\,\,\sin \,\,\pi \,t\,+\,4
(c) \displaystyle x\left( t \right)\,=\,\frac{1}{\pi }\,\,\sin \,\,\pi \,t\,+\,4
(d) None of these
Ans. (c)
9.
The velocity of a particle moving on the x-axis is given by v = x2 + x where v is in m/s and x is in m. Its acceleration in m/s2 when passing through the point x = 2m is
(a) 0
(b) 5
(c) 11
(d) 30
Ans. (d)
10.
The position of a particle moving along a straight line is given by x = 3sin(pt), where t is in sec and x is in m. Find the distance (in cm) travelled by the particle from t =1/4sec to t =3/4sec. \left( {\sqrt{2}=1.4} \right)
(a)175cm
(b)176cm
(c)172cm
(d)171cm
Ans. (d)
11.
The acceleration `a’ of a particle increases linearly with time t as a = 6t. If the initial velocity of the particle is zero and the particle starts from the origin, then the graph describing the motion of the particle is
(a) a parabola on velocity vs time graph
(b) a straight line on acceleration vs time graph
(c) a cubical curve on position vs time graph
(d) all of the above
Ans. (d)